mathematics and statistics - Te Kura

2012/1

te mĀtauranga pĀngarau te tauanga

mathematics and statistics

MX2062algebraic methods 2

ncea level 2

© te aho o te kura pounamu

mathematics and statistics ncea level 2

Expected time to complete work

This work will take you about 10 hours to complete.

You will work towards the following standard:

Achievement Standard 91261 (Version 1) Mathematics and Statistics 2.6

Apply algebraic methods in solving problems

Level 2, External

4 credits

Achievement Standard 91269 (Version 1) Mathematics and Statistics 2.14

Apply systems of equations in solving problems

Level 2, Internal

2 credits

In this booklet you will focus on these learning outcomes:

• manipulate algebraic expressions

• determine the nature of the roots of a quadratic equation

• forming, use and solve linear and quadratic equations

• solve simultaneous linear and non-linear equations.

You will continue to work towards Achievement Standard 91261 (2.6) in booklet MX2063.

You will continue to work towards Achievement Standard 91261 (2.14) in booklet

MX2063. The other work is covered in MX2021, MX2022 and MX2023.

Copyright © 2012 Board of Trustees of Te Aho o Te Kura Pounamu, Private Bag 39992, Wellington Mail Centre, Lower Hutt 5045,

New Zealand. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means without the

written permission of Te Aho o Te Kura Pounamu.

1MX2062 © te aho o te kura pounamu

contents

1 Equation-solving techniques

2 Translating words into mathematics

3 Quadratic equations

4 Problems involving quadratic equations

5 Problems involving simultaneous equations

6 Review activity

7 Answer guide

how to do the work

2 MX2062 © te aho o te kura pounamu

When you see:

Complete the activity.

Check your answers.

Your teacher will assess this work.

Use a calculator.

Contact your mathematics and statistics teacher.

You will need: • paper to work on, an exercise book with squared paper is recommended • a scientific or graphical calculator • access to OTLE, but this is not essential.

Resource overviewYou will need to write the answers on your own paper.

A graphics calculator is recommended but is not essential. Using a graphics calculator will enhance your understanding and is permitted in the assessment activities.

At the end of each exercise, mark your work using the Answer guide. The answers will give you useful feedback and help you in the learning process.

You should telephone or email your mathematics and statistics teacher if you would like to discuss your work.

At the completion of this booklet and your work on the review activity, complete the self-assessment page and the cover sheet, and send these to your mathematics and statistics teacher to assess. Your self-assessment will be returned to assist you with revising for these achievement standards.

1A


equation-solving techniques1

learning outcomesForm, use and solve linear and quadratic equations.Solve simultaneous linear and non-linear equations.

learning intentionsIn this lesson you will learn to:

• manipulate algebraic equations to solve • form, use and solve linear and quadratic equations • solve simultaneous equations by substitution, graphing and elimination methods.

introductionThis booklet is about using the concepts of algebra you have learnt in the first booklet and developing these ideas to solve problems.

You will learn how to translate words into mathematical symbols and how to set up equations to model situations.

You will use methods for solving linear and quadratic equations, and review how to solve simultaneous equations both linear and quadratic.

revision: linear and quadratic equationsSolve for x:

1. 2(x – 5) = 5x + 2

2. 3(3x + 2) – 2(x – 3) = 5

3. = – 6

4. x² – 5x + 6 = 0

5. x² = 3x

6. 3x² = 7x + 6

7. x + 12 = x²

Check your answers.

1A

x8

x2


equation-solving techniques

simultaneous equationsAll of the equations you have solved so far have had only one variable. Simultaneous equationsinvolve two variables. They are called simultaneous equations because a solution is found that satisfies two (or more) equations at the same time.

The simplest form of simultaneous equation is two linear equations; for example, y = 2x – 5 and 2x + 3y = 9. We shall deal with this type first.

There are three main methods of finding a solution: • graphing • elimination • substitution(wedealwiththisnextlesson).

graphingThe method here is to draw the graph of both equations on the same pair of axes and then read the x and y coordinates off the point of intersection.

Example 1Find the solution of the equations y = 2x – 5 and 2x + 3y = 9.The equation y = 2x – 5 is written in the form y = mx + c, so we can say directly that this line has:

gradient (m) = 2 and y-intercept (c) = –5

The equation 2x + 3y = 9 is more suited to the two-intercept method, because we can use x = 0 and y = 0 to find where the line cuts each axis.

If x = 0, 2 × 0 + 3y = 9

y = 3

If y = 0, 2x + 3 × 0 = 9

x = 4

Now draw both lines on one pair of axes and read off the point of intersection.

The lines meet at (3, 1), so the solution is x = 3, y = 1.

12

3

-5

0

y

x

y = 2x – 5

2x + 3y = 9



Solve these simultaneous equations using the graphing method.

1. y = 3x – 3 2. y = 2x + 6

y = –2x + 7 2y + x = 7

Check your answers.

eliminationThe second method for solving simultaneous equations is called elimination.

In this method, you first multiply each equation by a number that will give the coefficient of either the x terms or the y terms the same value but opposite signs. Then you add the two equations. This eliminates either the x terms or the y terms and makes it possible to find a value for the remaining variable.

Before starting this process, check that you can do all the necessary steps individually. Answer these questions then check your answers with those that follow.

revision1. What is the coefficient of y in 3x – 2y = 5?

2. Multiply both sides of these equations by 4.

a. 3x – 2y = 5

b. x + 4y = 2

c. –2x – y = 1

3. Multiply both sides of these equations by –2.

a. 2x + 5y = –11

b. –3x – 2y = 4

c. x + 4y = 7

Solutions:

1. –2

2. a. 12x – 8y = 20 b. 4x + 16y = 8 c. –8x – 4y = 4

3. a. –4x – 10y = 22 b. 6x + 4y = –8 c. –2x – 8y = –14

1B



Example 2Solve by elimination: x + 2y = 6 (A) x – y = 3 (B)

We will use a letter to represent each equation for reference.We could eliminate x or y, but this time we’ll eliminate y.

To eliminate y, these coefficients need to have the same absolute value, but the opposite sign. Multiplying equation (B) by 2 makes the coefficient of its y term equal –2. This has the same absolute value but the opposite sign to the coefficient of y in equation (A), as required.

Multiply equation (B) by 2: 2x – 2y = 6 (C)Rewrite equation (A) x + 2y = 6 (A)

Next, add equations (C) and (A) to eliminate y. 2x + x – 2y + 2y = 6 + 6 3x + 0y = 12 3x = 12 x = 4

Notice how the y terms were eliminated by adding.

Now find y by substituting 4 for x in equation (A) or (B).

In (A): x + 2y = 6 4 + 2y = 6 4 + 2y – 4 = 6 – 4 2y = 2 y = 1

The solution is (4, 1).

As you found y by substituting in equation (A), you should do your check from equation (B).

x – y = 4 – 1 = 3 This pair of equations could also be solved by multiplying equation (B) by –1 to produce coefficients for x of +1 and –1. You would then add the equations, eliminating x, and solve the resulting linear equation for y.



Example 3We will use the same equations in example 2 but this time we will eliminate x. x + 2y = 6 (A) x + y = 3 (B)Multiply (B) by –1: –x + y = –3 (C)Rewrite equation (A) x + 2y = 6 (A)Add (C) and (A) to eliminate x. 3y = 12 y = 1Now find x by substituting y = 1 into (A) or (B). x + 2(1) = 6 x + 2 = 6 x = 4

The solution is (4, 1).Check in (B) x – y = 4 – 1 = 3 ✓

Example 4Find the solution of these simultaneous equations: 3x – 2y + 4 = 0 (A) 2x + 5y + 28 = 0 (B)This time we shall eliminate x.Multiply (A) by 2: 6x – 4y + 8 = 0 (C)Multiply (B) by –3: –6x – 15y – 84 = 0 (D)Add to eliminate x: –19y – 76 = 0 –19y = 76 y = –4Substitute y = –4 into (B): 2x + 5 × –4 + 28 = 0 2x – 20 + 28 = 0 2x + 8 = 0 2x = –8 2x ÷ 2 = –8 ÷ 2 x = –4The solution is (–4, –4).Check in (A) 3x – 2y + 4 = 3 × –4 – 2 × –4 + 4 = –12 + 8 + 4 = 0 ✓



1. Solve this pair of simultaneous equations by elimination.

4x – 3y = 7 2x + 3y = –1

2. Solve by elimination. a. 7x – 2y = 3 b. 2x + y = 10 3x + 8y = –43 3x + 5y = 29

c. 2x – 3y = 1 d. 3x + 2y = 13 3x – 4y = 7 7x + 3y = 27

Check your answers.

substitutionThe aim of the substitution method is to develop one equation with only one variable. This can be x or y.

First, ensure that one of the equations is written in the form y = …Then substitute or replace the variable y in the other equation with the expression that y equals.

The resulting linear equation in x can then be solved. Finally, the value of y is found.

Example 5Find the solution of the simultaneous equations y = 3x – 2 y = x + 1

y = 3x – 2 (A) Number the equations for reference.

y = x + 1 (B)

x + 1 = 3x – 2

1 + 2 = 3x – x Collect like terms.

3 = 2x Solve for x.

x =

From (B): y = x + 1

so y =

y =

32

The y values are the same in each equation,so the y in equation (A) can be replacedwith x + 1 from equation (B).

Find the value of y by substituting the x value into either of the original equations.

3252

1C



It is a good idea to check your solution by substituting both values for x and y into the other original equation.

In (A), y = 3x – 2

= 3 × – 2

= as required.

The solution is x = , y = , which we may also write as ( , ).

Sometimes you may need to rearrange one of the equations before you can substitute.

Example 6Solve the following pair of simultaneous equations.

x + 3y = 11 (A) 3x + 2y = 12 (B)

Start by rearranging the first equation to obtain an expression for x.

x + 3y = 11 gives x = 11 – 3y

Continue by substituting 11 – 3y for x in equation (B). 3x + 2y = 12 (B) becomes 3(11 – 3y) + 2y = 12

Now solve for y: 33 – 9y + 2y = 12 33 – 7y = 12 33 – 7y – 33 = 12 – 33 –7y = –21 y = 3

Now substitute to find x:In (A): x + 3y = 11 x + 9 = 11 x = 2

Check your answers in equation (B): 3x + 2y = 12 3 × 2 + 2 × 3 = 12 6 + 6 = 12 12 = 12

The solution of the pair of equations is (2, 3).

32

5232

52

32

52



Solve the following pairs of simultaneous equations.

1. a. y = 4x + 1 (A) y = x + 4 (B)

Hint: Substitute x + 4 for y in (A).

b. y = 2x – 3 y = 4x + 5

2. a. 2x + 3y = 1 (A) y = 2x – 5 (B)

Hint: Substitute 2x – 5 for y in (A).

b. 2x – 3y = 11 (A) y = 3x – 6 (B)

c. 3x + 4y = 31 (A) 2x = y + 6 (B)

Hint: Rewrite equation (B) with y as the subject.

d. 6x – 3y = –8 x + 2y = 7

Check your answers.

Which method to use? The substitution method is generally the most useful. It can be applied regardless of the way the equations are written.

The elimination method is useful where neither of the coefficients of the x or y terms is 1.

The graphical method gives a very good visual of the linear equations and where the intersection point is.

1D


learning outcomeForm, use and solve linear and quadratic equations.


• change word problems into mathematical expressions or equations. • form and solve linear equations.

introduction If you were translating from English into a foreign language, you’d probably start with the words you know, then you’d put phrases together, and finally sentences.

English SpanishWords The cat El gatoPhrases On the mat En la esteraSentences The cat is sitting El gato está sentado on the mat en la estera

In a similar way, words, phrases and sentences can be translated into the language of mathematics.

English MathematicsWords Four 4Phrases Three plus four 3 + 4Sentences The sum of three 3 + 4 = 7 and four is seven

In this lesson, you will practise translating from English to Mathematics. We’ve assumed you know most of the words, but there are a couple of special cases where quite lengthy English phrases can be translated into a single mathematical term.

English Mathematics I’m thinking of a number x A certain number x The variable x

Here are some examples of phrases.

English Mathematics The sum of a and b a + b The product of seven and a 7a Twice as big as x 2x

One third of y y or

g is increased by four g + 4

translating words into mathematics 2

13

y3


translating words into mathematics

1. Start with the number x, and write down the results of each of the following operations.

a. Multiply it by two

b. Multiply it by minus one half

c. Divide it by four

d. Add three to it

e. Subtract five from it

f. Subtract it from five

g. Multiply it by two and then add three to the result

h. Add three to it and then multiply the result by two

i. Subtract five and then divide the result by two

j. Divide by seven and then add three to the result

2. Write these phrases using mathematical symbols.

a. seven more than x

b. The product of b and seven

c. six decreased by m

d. g multiplied by two

e. Double x and add two

f. Half of a

g. Three more than twice h

h. x decreased by four

i. One third of y

j. Five less than twice p

Check your answers.

2A

hx = xh



translating sentencesAs you do this section, concentrate on the translations. Do not solve the equations.

Cover the right-hand column of the table with a piece of paper. See if you can write each word sentence using mathematical symbols. Check each answer before attempting the next. We’ve used a for the variable in number 1, b in number 2, and so on.

Word sentence Mathematics

1. A number multiplied by five equals 10. 5a = 10 or 5x = 10 etc

2. Three more than a number is equal to 17. b + 3 = 17 or x + 3 = 17 etc

3. I think of a number and subtract three; the result is the same as multiplying the original number by four. c – 3 = 4c or p – 3 = 4p etc

4. I think of a number and square it; the result is 20 more than the original number. d² = d + 20 or x2 = x + 20 etc

1. Write each of the following sentences as a mathematical equation. Do not solve the equation.

a. If two is taken from z, the result is four.

b. The sum of q and six is 10.

c. If four is added to m, the result is nine.

d. The product of n and seven is 21.

e. If x is taken from 17, the result is four.

2B



2. Write each statement as an equation involving x. Do not solve the equation.

a. I think of a number, multiply it by three and add seven; the result is 42.

b. I think of a number, subtract five and then square the result. The final answer is 36.

c. I think of a number and add six; the result of doing that is the same as multiplying the original number by three.

d. I think of a number and add it to its reciprocal. The result is 4.25.

e. I think of a number, square it and then subtract it from twice the original number. The result is –32.

3. For each of the following:

i. Introduce and define two variables.

ii. Form two equations from the conditions in the problem. Do not solve the equation.

a. I’m thinking of two numbers. Their sum is 42 and their difference is three.

b. I’m thinking of two numbers. Their sum is 27 and their product is 50.

c. I’m thinking of two numbers. Their sum is four and the sum of their reciprocals is –0.3.

Check your answers.

Each of the translations you have done has been based on mathematical statements, where the words talk about numbers. The next few statements are more practical; they come from real-life situations. In each case, you will have to decide what you are asked to find, introduce a letter or variable for this, and state clearly what it represents. Again, concentrate on the translation; solving these equations will come later.



Example 1Write a mathematical sentence for the following problem.

Inthreeyearsʼtime,Billwillbe46.Howoldishenow?

SolutionYou are asked to find out how old Bill is now, so start by introducing a letter for this.

Write: Let x years be Bill’s age now.

Now translate the words into mathematical symbols: x + 3 = 46

By solving this equation, you could find Bill’s age, but that is not important at this stage.Concentrate on the process leading to finding the equation.

Example 2Translate into mathematics:

There were 18 dinner mints in a box. They were passed around at a dinner party and each person had two. There were four left. How many people were at the dinner party?

Solution

Introduce a letter Let the number of people be p.for the unknown quantity.

Think. If one person eats 2 dinner mints, then p people eat 2p dinner mints.

Translate. 2p + 4 = 18 The mints eaten the mints left the number in the box.

Solving this equation would tell you how many people were at the dinner party.



1. In the van Heusen family, Sonja is m years old. Complete each of the following statements about the ages of the other members of the family with mathematical symbols.

a. Pieter is twice as old as Sonja, so Pieter’s age is .

b. Hans is five years younger than Sonja, so Hans’s age is .

c. Wilhemina is three years older than Pieter, so Wilhemina’s age is .

d. Marita is two years younger than Hans, so Marita’s age is .

2. For each of the following problems:

i. Introduce a letter and state clearly what it represents.

ii. Write an equation (but do not solve it). a. The distance from Upper Hutt to Wellington is 34 km. Seth has driven 15 km. How far has he still to go? b. In a test, the top mark was four times as high as the bottom mark. If the top mark was 48, what was the bottom mark? c. There were 24 pieces in a block of chocolate. The children each had three pieces and there were three pieces left over. How many children shared the chocolate? d. Joan had a quick way of remembering the postal code for her Australian pen friend. It is five more than five times 500. What is the postal code? e. The length of a rectangular lawn is 18 m. If the perimeter of the lawn is 56 m, how wide is it? f. At a sheep sale, George sold 50 more sheep than Charlie. Together they sold 160 sheep. How many sheep did Charlie sell?

Check your answers.

2C



problems involving linear equations Some people have an intuitive approach to problem solving; others work by trial and error. However,asystematic,step-by-stepapproachisusuallymuchmoreefficientandmuchmorelikely to lead to a successful solution.

We recommend that you use these steps in all of the following problems, even if you can see the answer straight away. By practising the routine on simple problems, you will develop the confidencetouseitonmoredifficultproblems.

The steps are: 1. Read through the problem carefully. Try restating it in your own words. 2. Consider drawing a diagram. 3. Choose a letter for the unknown number or quantity and state clearly what it represents. 4. Form an equation from the conditions in the problem. 5. Solve the equation. 6. Check your answer with the conditions stated in the problem.

7. Write an English sentence answering the original question.

Example 3A rectangle is three times as long as it is wide. Its perimeter is 32 cm. What is its width?

Solution1. Read through the problem carefully. Think: The length is three times the width, and the distance right around is 32 cm.

2. Consider drawing a diagram. Can you draw one that illustrates this problem? Yes. This will be useful in step 4.



3. Choose a letter for the unknown number and state clearly what it represents. It’susuallyagoodideatolettheletterstandforthenumberorquantityyouaretryingtofind. In this case you’d write: Let the width of the rectangle be x cm.

4. Form an equation from the conditions in the problem. This is where your diagram can help.

The width is x. The length is ‘three times as longasitiswideʼ.

length = 3x

Theperimeteris32cm.Tofindtheperimeter,addthelengthsofthefoursides.

The equation is x + 3x + x + 3x = 32.

5. Solve the equation. x + 3x + x + 3x = 32 8x = 32 x = 4

6. Check your answer with the conditions stated in the problem. The width is 4 cm The length is 3 × 4 = 12 cm The perimeter is 4 + 12 + 4 + 12 = 32 cm

7. Write a sentence answering the original question. The width of the rectangle is 4 cm.

x x

3x

3x



Example 4Keri paid $7.75 when she bought a roll, some fruit and a can of juice for lunch. She paid $3.25 more for the fruit than the roll and $1.75 more for the roll than the juice. How much did the roll cost?

You may like to try solving this problem yourself using the guidelines shown in Example 3. If not, work through the following questions. This will help you learn the problem-solving process.

1.Whatareyouaskedtofind?

2. Choose a letter for the unknown number and state clearly what it represents.

3. Write a mathematical statement for the price of the fruit.

4. Write a mathematical statement for the price of the juice.

5. Write an equation that connects the separate prices of the roll, fruit and juice, and the total price.

6. Solve your equation in (5).

7. a. What is the price of the roll?

b. What is the price of the fruit?

c. What is the price of the juice?

d. What is the total cost of the lunch?

e. Do these answers agree with the information in the question?

8. Write a sentence answering the original problem.

Answer to example 4

1. The price of the roll.

2. Let the price of the roll be r $.

3. Price of fruit = r – 3.25

4. Price of juice = r – 1.75

5. r + (r – 3.25) + (r – 1.75) = 7.75

6. 3r – 5 = 7.75

3r = 12.75

r = 4.25



7. a. The roll costs $4.25

b. The fruit costs $1.00

c. The juice costs $2.50

d. The total cost is $7.75

e. Yes.

8. The cost of the roll is $4.25

In practice, you’d write solutions like this.

Example 5The sum of four consecutive integers is 3 810. Find the integers.

Let prepresentthefirstinteger. p + (p + 1) + (p + 2) + (p + 3) = 3 810 p + p + 1 + p + 2 + p + 3 = 3 810 4p + 6 = 3 810 4p = 3 804 p = 951

So p + 1 = 952 p + 2 = 953 p + 3 = 954

Check 951 + 952 + 953 + 954 = 3810

The four consecutive integers that add to 3810 are 951, 952, 953 and 954.

Consecutive means they come one after the other. 5, 6, 7 and 8 are four consecutive integers.



2D Inthefollowingproblems,allvariable(s)mustbedefinedappropriately.

1. A rectangular lawn is twice as long as it is wide. Its perimeter is 96 m. What is its width?

2. Thesumoffiveconsecutivewholenumbersis725. Whatisthefirstnumber?

3. Ken paid $8.80 when he bought a pie, a cake and a bottle of cola for lunch. If the cake cost 40 c more than the cola, and the pie cost twice as much as the cake, how much was the cake?

4. Amotoristtravelsatacertainspeedfortwohours,andthen,afterbeingfinedforspeeding, does the next three hours at 15 k/m less. Ifthetotaldistancetravelledis480kilometres,whatwashisspeedforthefirsttwohours?

If you have trouble doing this question by yourself, work through the following steps:

1. Choose a letter for the unknown value and state what it represents. 2. Writeamathematicalstatementforthedistancetravelledinthefirsttwohours. (Hint: Distance = speed × time) 3. Write a mathematical statement for the speed at which the motorist was travelling after he got his ticket. 4. Write a statement for the distance travelled in the last three hours. 5. Write an equation for the total distance travelled. 6. Solve the equation. 7. Write a sentence answering the original question. 8. Check your answer.



5. One tree is seven times as old as another. Five years from now, it will be four times as old as the younger tree is then. How old is the younger tree now? Hint: a. If the younger tree is x years old now: i. how old is the older tree now? ii. howoldistheyoungertreeinfiveyearsʼtime? iii.howoldistheoldertreeinfiveyearsʼtime? b. Writeanequationconnectingtheagesofthetreesinfiveyearsʼtime. c. Solvetheequation.Checkyouranswerandwriteafinalsentence.

6. Supplementary angles add to 180°. If one of two supplementary angles is 3º smaller than twice the other, what is the size of the smaller angle?

7. Several years ago Jeremy Jones joined a club. The entrance fee was $35. Theyearlysubscriptionforthefirstthreeyearswas$240andafterthatitroseto$336. He calculated that he’d spent a total of $2 099. How many years ago did he join the club?

Check your answers.


quadratic equations3

learning outcomesForm, use and solve linear and quadratic equations.Determine the nature of the roots of a quadratic equation.


• form and solve quadratic equations using the quadratic formula • determine the nature of the roots of a quadratic equation using the discriminant.

introduction Many problems involve other types of equations. In this lesson, you’ll look at quadratic equations and how to solve them using the quadratic formula.

2x² + x – 3 = 0 and x² + 5x + 6 = 0 are examples of quadratic equations.

The method you have probably used to solve quadratic equations so far is based on factorising (using brackets).

Example 1Solve x² + 5x + 6 = 0

(x + 3)(x + 2) = 0Either x + 3 = 0 or x + 2 = 0 x = –3 or x = –2The solutions are x = −3 and x = −2.

Example 2Solve 2x² + x – 3 = 0 2x² + x – 3 = 0

(2x + 3)(x – 1) = 0Either 2x + 3 = 0 or x – 1 = 0 2x = –3 or x = 1

x = –

The solutions are x = − and x = 1.

In real life problems, many quadratic equations cannot be factorised.

323

2


quadratic equations

A useful formula helps deal with these cases.

For any quadratic equation in the form ax² + bx + c = 0

then x =

Thesymbol±isreadas‘plusorminusʼ.Youcanwritethetwosolutionsas:

x = or x =

You can use this formula for any quadratic equation, but for those that are easy to factorise,you may find factorising a quicker method.

If you’d like to see how this formula is formed, look in the appendix that is included after the Answer guide.

remember! If ax² + bx + c = 0

then x =

Example 3

Solve 2x² + 6x + 3 = 0

ax² + bx + c = 0

a = 2, b = 6, c = 3

x =

x =

x =

x =

The two solutions are:

x = or x = x = −0.6339746 or x = −2.3660254

−b ± b²−4ac2a

−b ± b²−4ac2a

−6 ± 6²−4 × 2 × 32 × 2

−6 ± 36−244

−6 ± 124

−6 + 124

−6 − 124

Compare the equationwith ax2 + bx + c = 0to find a, b and c.

Write the formula.

Subtitute the values for a, b and c.

−b ± b²−4ac2a

−b + b²−4ac2a

−b − b²−4ac2a


quadratic equations

Round your solutions sensibly to put them in the context of the problem.

x = –0.63 or x = –2.37 (2 d.p.)

Follow the next examples through before attemping activity 3A.Use your calculator to find decimal approximations for the solutions.

Example 4Solve x² – 3x + 1 = 0

a = 1, b = –3, c = 1 Write out the values for a, b and c.

x = Write the formula.

x = Substitute in a, b and c.

x = Simplify.

x = or x =

x = 2.62 or x = 0.38 (2 d.p.) Any variable can be used instead of x. Just change x in the formula to a different variable.

Example 5Solve 2p² + 5p = 2 2p² + 5p – 2 = 0 Move all terms to the left of the = sign.

a = 2, b = 5, c = –2 Write out the values for a, b and c.

p = Write the formula.

p = Substitute a, b and c.

p = Simplify.

p =

p = or p =

p = 0.35 or p = –2.85 (2 d.p.)

Now try the following exercise to practise using the quadratic formula. Note: The solutions of quadratic equations are frequently called roots.

−b ± b²−4ac2a

−(−3) ± (−3)²−4×1×12 × 1

3 ± 52

3 + 52

3 − 52

−b ± b²−4ac2a

−5 ± 5²−4 × 2 × −22 × 2

−5 ± 25 + 164

−5 ± 414

−5 + 414

−5 − 414


quadratic equations

Solve these equations using the formula.Remember to check that all terms are on the left-hand side of the equals sign first.If not, rearrange the equation, setting the right-hand side to zero, as in the last example.

1. x² + 5x + 3 = 0 2. 4x² + 9x – 4 = 0

3. x² – 3x – 2 = 0 4. 7s² – 3s = 1

5. x² – 6x + 5 = 0 6. 4x² = –12x – 9

Check your answers.

the nature of the roots of a quadratic equationThe solutions to quadratic equations are sometimes called roots.For example, the roots of (x – 3)(x + 2) = 0 are 3 and –2.

As you will see in this booklet, it is often useful to know what the roots of a quadratic equation are like without having to solve the equation. This is called finding or describing the nature of the roots.

If you solved lots and lots of quadratic equations, you would find that the equations could be divided into four different types, grouped according to the nature of their roots.

1. Equations with two irrational roots; for example, 7s² – 3s = 1

s =

37 is a positive number, so it has a square root. There are two real roots:

s = and s =

But 37 cannot be found exactly. 37 is a surd, and so the two roots are irrational.

2. Equations with two rational roots; for example, x² – 6x + 5 = 0

x =

16 is a positive number, so it has a square root. There are two real roots:

x = and x =

Now 16 = 4, so the roots can be found exactly:

x = and x =

x = 5 and x = 1

So the two roots are rational.

3A

3 ± 3714

3 − 3714

3 + 3714

6 ± 162

6 − 162

6 + 162

6 + 42

6−42


quadratic equations

-12 ± 08

−128

1 ± −116

3. Equations with two equal rational roots; for example, 4x² + 12x + 9 = 0

x =

0 = 0, so there is only one value for x.

x =

= –1.5

So there are two equal rational roots. Alternatively, you can think of this equation as having one repeated root or two equal roots.

4. Equations with no real roots; for example, 3y² = y – 1

y =

–11 is less than zero, so it has no real square root. y cannot be calculated in this case. There are no real roots.

the discriminantThe part of the formula that determines the nature of the roots is the expression under the square root sign, b² – 4ac. Because b² – 4ac distinguishes, or discriminates, between different kinds of roots, it is called the discriminant of the quadratic equation ax² + bx + c = 0.

the discriminant of a quadratic equation For the equation ax² + bx + c = 0

the quadratic formula is x =

The discriminant b² – 4ac is the part of the formula under the sign.

You will use the discriminant to find the number of real solutions of a quadratic equation ax² + bx + c = 0.

b² – 4ac > 0 When the number under the sign in the quadratic formula is positive, there will be two distinct real solutions:

x = or x =

b² – 4ac = 0 When the square root part of the formula is zero, there will be two real and equal solutions: x =

b² – 4ac < 0 Here, the number under the sign in the formula is negative. It is not possible to find the square root of a negative number, so there are no real solutions.

−b ± b²−4ac2a

−b − b²−4ac2a

−b + b²−4ac2a

−b2a


quadratic equations

Example 6Find the nature of the roots of x² + 7x + 10 = 0.SolutionFirst, compare with the general equation to find a, b and c.

x² + 7x + 10 = 0

ax² + bx + c = 0

a = 1, b = 7, c = 10

Now find the value of b² – 4ac.

b² – 4ac = 7² – 4 × 1 × 10 = 49 – 40 = 9b² – 4ac > 0, so there are two roots.

Nine is the square of a rational number, so the roots are rational. Therefore, x² + 7x + 10 = 0 has two rational roots.

Example 7Find the value of the discriminant of 6x² + 2x + 5 = 0 and state the number of real solutions for the equation.

a = 6, b = 2, c = 5

The discriminant b² – 4ac = 2² – 4 × 6 × 5

= –116

b² – 4ac < 0, so there are no real solutions.

1. The discriminant of 9x² – 2x + 1 = 0 is –32. What is the nature of the roots?

2. The value of b² – 4ac for a particular quadratic equation is 0. Describe the roots.

3 The discriminant of 2x² – 8x – 1 = 0 is 72. What information does this give about the nature of the roots?

4. Find the nature of the roots of each of the following equations.

a. x² + 3x – 10 = 0

b. x² + 3x – 11 = 0

c. 4x² – 4x + 1 = 0

d. x² + 2x + 1 = 0

e. 3x² – 2x + 3 = 0

f. 2x² – 5x + 2 = 0

3B


quadratic equations

5. For what values of m will the roots of x² – 6x + m = 0 be real?

6. The equation 2x² + kx + 18 = 0 has equal roots. Find two possible values of k.

Check your answers.

remember! The discriminant b² – 4ac determines the nature of the roots of ax² + bx + c = 0.

1. If b² – 4ac > 0, there are two real roots. If b² – 4ac = m², where m Q, the roots are rational. Otherwise, they are irrational.

2. If b² – 4ac = 0, there is one rational root.

3. If b² – 4ac < 0, there are no real roots.

Forwhichequationsfromthelist1–7couldyoufind:

a. two distinct solutions?

b. two equal solutions?

c. no real solutions?

1. x² + 5x + 3 = 0

2. 4x² + 9x – 4 = 0

3. x² – 3x – 2 = 0

4. 7x² – 3x = 1

5. x² – 6x + 5 = 0

6. 4x² + 12x + 9 = 0

7. 3y² = y – 1

8. Which equations have real solutions?

9. Which equation has no real solution?

10. Which equations have rational solutions?

11. Which equations have irrational solutions?

Check your answers.

3C


quadratic equations

discriminants and quadratic graphsYou will now learn how the graphs of quadratic functions illustrate the solutions to quadratic equations. This work combines the following topics:

• the solutions to quadratic equations (use of quadratic equation)• the nature of the roots (the discriminant)• the graphs of quadratic functions (graphing).

discovery exercise1. Answer the following questions, for the function f(x) = x² – 4x + 3 = (x – 1)(x – 3)

a. For the general quadratic equation ax² + bx + c = 0 the discriminant is given by b² – 4ac. Find the discriminant of the equation f(x) = 0.

b. Solve the quadratic equation f(x) = 0.

c. Give the point(s), if any, where the graph of f intersects the x-axis.

d. Sketch the graph of the function f.

Check your answers.

2. Answer the following questions, for the function g(x) = x² – 4x + 4 = (x – 2)²

a. Find the discriminant of g(x) = 0.

b. Solve g(x) = 0.

c. Give the point(s), if any, where the graph of g intersects the x-axis.

d. Sketch the graph of g.

3. Answer the following questions, for the function h(x) = x² – 4x + 5

a. Find the discriminant of h(x) = 0.

b. Solve h(x) = 0.

c. Give the point(s), if any, where the graph of h intersects the x-axis.

d. Sketch the graph of h.


quadratic equations

4. Copy and complete the table from your answers to questions 1, 2 and 3.

FunctionIntersection point(s) on

x-axis

Sign of b² – 4ac (+, 0, -)

Solutions of y = 0

Number of real roots of y = 0

f(x)

g (x)

h (x)

(1, 0), (3, 0) + x {1, 3} 2

Look for patterns in the table above.

• Whereonthegraphwouldyoufindthesolutiontoy = 0?

• What is the connection between the number of real roots and the number of times the parabola cuts the x-axis?

• What is the connection between the discriminant and the number of times the parabola cuts the x-axis?

Copyandcompletethefollowingstatementstosummariseyourfindings.

a. The roots of the equation ax² + bx + c = 0 are the of the points where

the graph of the function y = ax² + bx + c intersects the .

b. i. If the graph of y = ax² + bx + c crosses the x-axis twice, then ax² + bx + c = 0

has real roots and b² – 4ac 0.

ii. If the graph of y = ax² + bx + c touches the x-axis, then ax² + bx + c = 0


iii. If the graph of y = ax² + bx + c does not intersect the x-axis, then ax² + bx + c = 0


Check your answers.


quadratic equations

remember! The points where the graph of y = ax² + bx + c crosses the x-axis are the points where y = 0; that is, where ax² + bx + c = 0. So the x-values of these points are the roots of the equation ax² + bx + c = 0.

If the graph crosses the x-axis twice, then ax² + bx + c = 0 has two real unequal roots; that is, the x-values of the two points where the graph crosses the x-axis. This occurs when b² – 4ac > 0.

If the graph touches the x-axis, then ax² + bx + c = 0 has twoequal real roots; that is, the x-value of the point where the graph touches the x-axis. (This is sometimes called a repeated root.) This occurs when b² – 4ac = 0.

If the graph does not cross the x-axis, then ax² + bx + c = 0 has no real roots. This occurs when b² – 4ac < 0.

y

0 x

y

0 x

y

0 x


quadratic equations

1. This graph shows a quadratic function f.

a. Use the graph to solve f(x) = 0.

b. Is the discriminant of f(x) = 0 positive, negative or zero?

2. The graph shows a quadratic function g.

a. What is the solution set of g(x) = 0?

b. Describe the roots of g(x) = 0.

c. What is the sign of the discriminant of g(x) = 0?

multichoiceWrite down the letter that corresponds to the answer you think is correct.

3. Which best describes the nature of the roots of 3x² – 4x + 2 = 0? a. The roots are not real. b. There are two distinct real roots. c. There is only one real root. d. Both roots are irrational. e. There are two distinct rational roots.

Question 4 refers to the following graphs.

I. II. III. IV.

4. The equation f (x) = 0 has two real roots for which functions? a. I only b. IV only c. II and III d. I and IV

Check your answers.

f (x)

0 x3-1

-3

x0

g(x)

3

0 4

f (x)

x

0 x-4

-3

f (x)

0 x-4

3

f (x)

3

f (x)

x40

3D


problems involving quadratic equations

learning outcomesForm, use and solve linear and quadratic equations.Determine the nature of the roots of a quadratic equation.


• form and solve quadratic equations using the quadratic formula • determine the nature of the roots of a quadratic equation and apply the discriminant

to find solutions. introductionSome algebraic problems need to be solved using quadratic equations. It is important that you check your answers with the conditions stated in the problem, as quadratic equations always have two solutions and one of these may be nonsensical. A negative length for example, or a negative age or a fraction of a person, does not make much sense.

Example 1Twice the square of a certain whole number is 91 more than 19 times that certain whole number. Find the number.

Let k be that number.

2k² = 19k + 91

2k² – 19k – 91 = 0 (2k + 7)(k – 13) = 0Either 2k + 7 = 0 or k – 13 = 0

k = or k = 13

Since the original number was a whole number, k = is not a solution to the problem.

Check If k = 13 2k² = 2 × 13² = 338 19k + 91 = 91 × 13 + 91 = 338

The number required is 13.

4

−72

Twice the square of the number.

19 times the original number.

91 more than.

−72


problems involving linear equations

Example 2A rectangular lawn measures 14 metres by 11 metres. A path of uniform width surrounds it. If the area of the path is 186 m2,findthewidthofthepath.

A diagram shows the problem more clearly.

Let the width of the path be x metres.

The area of the path is the difference between the areas of the outer and inner rectangles.

Area of outer rectangle = (11 + 2x)(14 + 2x)

Area of inner rectangle = 11 × 14

(2x + 11)(2x + 14) – 11 × 14 = 186

4x² + 50x + 154 – 154 = 186

4x² + 50x – 186 = 0

2x² + 25x – 93 = 0

(2x + 31)(x – 3) = 0

Either x = or x = 3

Since x = –15.5 is not a realistic answer, the required answer is 3.

Check Area of outer rectangle = 17 × 20 = 340 Area of inner rectangle = 154 Area of path = 340 – 154 = 186The width of the path is three metres.

14

11

−312

Make it simpler bydividing throughby a common factor.

xx


problems involving linear equations

For some of these problems, you may need to use the quadratic formula.

x =

1. The temperature T, in degrees centigrade, of the element of a toaster is given by T = 18 + 70x – 4x² (0 < x < 8) where x is the time in seconds since the toaster is turned on. Howlongdoesittakeforthetemperatureoftheelementtofirstreach300°C?

2. The vertical height h, in metres, of a ball thrown upwards by a boy is given by h = 2 + 20t – 5t² where t is the time in seconds since the ball left the boy’s hand. At what times is the ball 18 metres above the ground? Explain why there are two different times.

3. A path of uniform width surrounds a rectangular lawn that measures 25 metres by 16 metres. If the area of the path is 180 m2,finditswidth.

4. A man was 26 years old when his son was born, and the product of their ages now is 192. How old is the son now?

5. Kim is four years older than Kit. The sum of the squares of their ages is 346. Find Kit’s age.

6. The lengths of three sides of a triangle in centimetres are x + 7, x – 2 and 3x, as shown in this diagram.

a. Use the Theorem of Pythagoras to findanequationlinking the lengths of the sides.

b. Solve the equation for x.

c. Which value for x is the correct solution?

d. What are the lengths of each side of the triangle?

Check your answers.

4A

−b ± b²−4ac2a

x−2

3x

x + 7


problems involving simultaneous equations

learning outcomeSolve simultaneous linear and non-linear equations.


• form, use and solve linear and quadratic equations • solve simultaneous equations.

introductionThe problems you will encounter in this lesson are solved using simultaneous equations. You will need to find two linear equations to describe each situation, as there are two unknowns in each situation. This means you will have to introduce two variables and state what each represents. Then follow the basic problem-solving techniques that you’ve used in the last two lessons.

Example 1The sum of two numbers is 23 and their difference is 27. What are the two numbers?

1. Read the problem carefully. Think:‘sumʼmeansaddingand‛differenceʼmeanssubtracting.

2. Consider drawing a diagram. There’s no point this time.

3. Choose letters for both unknown numbers and state what they represent. Let the numbers be a and b.

4. Form two equations from the information in the question. a + b = 23 The sum is 23 a – b = 27 The difference is 27

5. Solve the pair of equations simultaneously. a + b = 23 (A) a – b = 27 (B) a + a + b – b = 23 + 27 Adding (A) and (B) 2a = 50 a = 25 25 + b = 23 Substitute into (B). b = –2

6. Check your answer with the conditions stated in the problem. 25 + –2 = 23 25 – –2 = 27

7. Write a sentence answering the question. The numbers are 25 and –2.

5



In example 1, you did not have to distinguish between a and b because you were finding two numbers. In example 2, it is important to state clearly what each variable represents.

Example 2

Concert tickets cost $110 for the main floor and $165 for the balcony. 2 000 tickets were sold. The takings were $260 700. How many balcony tickets were sold and how many main-floor tickets were sold?

You may like to try solving this problem yourself using the guidelines shown in example 1. If not, work through the following questions, which will help you learn the problem-solving process.

1. What are you asked to find?

2. Choose a letter to represent the number of balcony seats sold and write a sentence stating this. Repeat for the number of main-floor tickets.

3. Write a mathematical phrase for the takings from balcony tickets. Write a mathematical phrase for the takings from main-floor tickets. Write an equation that connects the statements for the takings of the balcony tickets and the main-floor tickets with the total takings.

4. What information in the question have you not yet used?

5. Write an equation connecting the number of balcony tickets and main-floor tickets sold with the total number of tickets sold.

6. Solve simultaneously the two equations you have found.

7. Write a sentence answering the original question.

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Answer to example 2

1. The number of balcony tickets and the number of main-floor tickets sold.

2. Let x represent the number of balcony tickets sold. Let y represent the number of main-floor tickets sold.

3. Balcony ticket takings: 165x Main-floor ticket takings: 110y 165x + 110y = 260 700

4. 2 000 tickets were sold.

5. x + y = 2 000

6. 165x + 110y = 260 700 (A) x + y = 2 000 (B) From (B) y = 2 000 – x Substitute 2 000 – x for y in (A) 165x + 110(2 000 – x) = 260 700 165x + 220 000 – 110x = 260 700 55x + 220 000 = 260 700 x = 740

Put 740 for x in (B) 740 + y = 2 000 y = 1 260

Check x + y = 2 000 1260 + 740 = 2 000 (Number of tickets) 165x + 110y = 260 700 165 × 740 + 110 × 1 260 = 260 700 (Takings)

7. Answer: 740 balcony tickets and 1 260 main-floor tickets were sold.



Use a pair of simultaneous equations to solve each of the following problems.

1. Find two numbers such that three times the first added to twice the second makes 30, and twice the first added to the second makes 17.

2. I’m thinking of two numbers. They differ by seven, and if I add twice the greater number to five times the smaller number I get 42.

3. Half the sum of two numbers is – , while half their difference is . Find the numbers.

4. A man was 32 years old when his son was born and the sum of their ages is now 56. How old are they both now?

5. Thirty-seven teenagers are going on a picnic. If there are nine more boys than girls, how many boys and how many girls are there?

6. Kelly weighs 3 kg less than Terri, and together their weights are still 5 kg less than Joe’s. If Joe weighs 90 kg, what does Kelly weigh and what does Terri weigh?

7. Ian paid $7.15 for five nectarines and three peaches. Hemi paid $8.60 for four nectarines and six peaches. What is the average price of a nectarine and a peach?

Check your answers.

5A

Remember! The variable(s) you define must correctly express what you’re trying to find.

12

32

iSto

ck

iSto

ck



simultaneous equations: one linear, one non-linearThe pairs of simultaneous equations you have solved so far have contained two linear equations, and you could use either the elimination or substitution methods to solve them algebraically.

For solving pairs of simultaneous equations where only one equation is linear, you must use substitution. You substitute an expression for x or y obtained from the linear equation into the non-linear equation.

The non-linear equation can take a variety of forms; for example, y = x² + 5x + 6 andxy = 12, x² + y² = 4, but the equation you obtain after substitution will be a quadratic equation.You can solve this by either factorising, or by using the quadratic formula.

Example 3Solve simultaneously y = x² (A) y = x + 2 (B)

Start by substituting x + 2 for y in equation (A). x + 2 = x²

Look closely at this equation. It contains an x² term, so it is a quadratic equation. Thefirststepinsolvingquadraticequationsistomovealltermsontothesameside. 0 = x² – x – 2 x² – x – 2 = 0

Nowfactorisetheleft-handsideandfindvaluesforx. (x – 2)(x + 1) = 0

Either x – 2 = 0 or x + 1 = 0 x = 2 x = –1

Continue by substituting each xvalueintoequation(B)tofindthey values.x = 2 gives y = 2 + 2 and x = –1 gives y = –1 + 2 = 4 = 1

Check your solutions in equation (A):4 = 2² so (2, 4) is a solution1 = (–1)² so (–1, 1) is a solution.

The solutions are: (x, y) {(2, 4), (–1, 1)}



Example 4Solve these simultaneous equations. x² + y² = 13 (A) x – y + 1 = 0 (B) x = y – 1 (C) Make x the subject of the formula in (B).

(y – 1)² + y² = 13 Substitute for x in (A).

2y² – 2y – 12 = 0 Simplify.

y² – y – 6 = 0 Divide by common factor 2.

(y – 3)(y + 2) = 0 Factorise.

Either y−3 = 0 or y + 2 = 0 Solve.

y = 3 y = −2

If y = 3, then x = 3 – 1 Substituteinto(C)tofind x.

= 2

If y = –2, then x = –2 – 1 = –3

Check in (A): 2² + 3² = 4 + 9 Check both answers using other equation. = 13 (–3)² + (–2)² = 9 + 4 = 13

Solution is: (x, y) {(2, 3), (–3, –2)} Write out solution.



Solve the following pairs of simultaneous equations algebraically.

1. y = x² + 3x – 2 y = x – 3

2. x² + y² = 10 y = 3x + 10

3. 2x – y + 1 = 0 y = x² + x – 5

4. x² + y² = 25 x + y = 7

5. xy = 15 4x – y = 7 Check your answers.

5B



problems involving simultaneous equations: one linear, one non-linearTo solve these problems, you will use the same basic techniques that you used earlier. Astherearetwounknownsineachproblem,youwillneedtodefineandusetwovariables.

Example 5The difference of two positive numbers is three and the sum of their squares is 65. What are the numbers?

1. Read the problem carefully. Think:‛differenceʼmeanssubtractand‛sumʼmeansadd. 2. Consider drawing a diagram. There’s no point this time.

3. Choose letters for both unknown numbers and state what they represent. Let the numbers be a and b.

4. Form two equations from the information in the question. a – b = 3 The difference is 3 a² + b² = 65 The sum of the squares is 65

5. Solve the equations simultaneously. a² + b² = 65 (A) a – b = 3 (B)

From (B): a = b + 3In (A): (b + 3)² + b² = 65 b² + 6b + 9 + b² = 65 2b² + 6b – 56 = 0 b² + 3b – 28 = 0 (b + 7)(b – 4) = 0Either b + 7 = 0 or b – 4 = 0 b = –7 b = 4

If b = –7, then a = –7 + 3 = −4

If b = 4, then a = 4 + 3 = 7



6. Check your solutions with the conditions stated in the problem. The question says the numbers are positive, so the negative answers must be eliminated.

7 – 4 = 3 7² + 4² = 49 + 16 = 65

7. Write a sentence answering the question. The two positive numbers are 7 and 4.

Example 6A rectangular paddock has a perimeter of 60 metres and an area of 216 m2. Find the paddock’s dimensions.

1. Read the problem carefully. Thinkhowyoufindtheareaandperimeterofrectangles.

2. Consider drawing a diagram.

3. Choose letters for both unknowns and state what they represent. Let the length of the paddock be x metres and let the width be y metres.

4. Form two equations from the information in the question 2(x + y) = 60 The perimeter is 60 m. xy = 216 The area is 216 m2.

y

x



5. Solve the equations simultaneously. 2(x + y) = 60 (A) xy = 216 (B) From (A): x + y = 30 y = 30 – x

In (B): x(30 – x) = 216 30x – x² = 216 x² – 30x + 216 = 0 (x – 12)(x – 18) = 0

Either x – 12 = 0 or x – 18 = 0 x = 12 x = 18

If x = 12, y = 30 – 12 = 18

If x = 18, y = 30 – 18 = 12

6. Check the solutions with the conditions stated in the problem.

Area = 18 × 12 = 216

Perimeter = 2(18 + 12) = 2(30) = 60

7. Write a sentence answering the question. The dimensions of the paddock are 18 metres by 12 metres.

12

18

Don't forget to include the units.



Use a pair of simultaneous equations to solve each of the following problems.

1. The sum of two numbers is 25 and their product is 144. What are the numbers?

2. Thedifferenceoftwopositivenumbersisfiveandthesumoftheirsquaresis97. Find the numbers.

3. A rectangle has a perimeter of 76 centimetres and an area of 325 cm². Find its dimensions.

4. The hypotenuse of a right-angled triangle is 51 centimetres long and the perimeter of the triangle is 120 centimetres. Find the lengths of the two shorter sides.

Check your answers.

points of intersection of lines with curvesYoucanfindorillustratethesolutionstosimultaneousequationslikethoseyousolvedinthelast lesson with graphs. The linear equations will produce a straight line; the non-linear equations a curve.

The coordinates of the point (or points) of intersection of the line and the curve are the solutions to the simultaneous equations being solved.

The solutions to two pairs of simultaneous equations are illustrated below.

1. Use the graphs to write down: i. the pair of simultaneous equations involved. ii. the solution of the pair of equations.

a. b.

2. Verify your answers to question 1 (ii) by solving the equations algebraically.

Check your answers.

5C

5D

y

0 1 x

3

y = (x + 3)(1 − x)y = 3

−1−2−3

0

y

x

x² + y² = 25

y = −2x + 5−3

3

5

54


learning outcomesManipulate algebraic expressions.Determine the nature of the roots of a quadratic equation.Form, use and solve linear and quadratic equations.Solve simultaneous linear and non-linear equations.


• review work on linear and quadratic equations • review work on simultaneous equations • answer questions for your teacher to assess.

1. Solve these pairs of simultaneous equations: a. x – 2y = 7 b. 2x + 5y = 34 y = 2x + 22 3x + 4y = 30

2. Express each of the following using algebraic symbols. a. Two more than five times a certain number. b. Six less than twice a certain number. c. The mean of two numbers. d. The number of minutes in a certain number of hours.

3. Find two consecutive even integers such that twice the smaller exceeds the larger by 18.

4. A Parent Teacher Association knows it sold a total of 300 tickets to a picnic and that the proceeds from the sale of the tickets was $1 080. Unfortunately, no record was kept of the separate number of adults’ and children's tickets sold. This information is now needed because a special treat is planned for the children. If the adults’ tickets cost $5 each and children’s tickets cost $3 each, solve the PTA’s problem.

5. The hypotenuse of a right-angled triangle is 34 cm long. Find the lengths of the other two sides if one side is 15 cm longer than the other.

6. Solve the following pair of simultaneous equations algebraically. x² + y² = 4 y = x – 2

7. Find the coordinates of the points of intersection of y = –5x + 1 and xy = –6.

8. Show that the line x + 3 = 0 is a tangent to the curve x² + y² = 9.

review activity6

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6A


review activity

9. The graph shows that the line y = 12 – x cuts the parabola y = x².

a. What pair of simultaneous equations could be solved using this information? b. Write down the solution to your equations in (a).

c. What quadratic equation could be solved using this information?

10. A piece of wire 56 centimetres long is bent to form a right-angled triangle with a hypotenuse 25 centimetres long. Find the lengths of the other two sides.

Check your answers.

When you check your answers to this review activity, you may discover that, although your answers, were correct, you used a different method. Many problems can be solved in more than one way. We’ve shown only one, and it’s not necessarily the best or easiest method.

y

x1 2 3-4 -2

1

2

3

56

7

8

9

10

11

12

13

14

15

16

y = 12 − xy = x²


review activity

This must be your own work.You must sign the cover sheet to confirm this.This is an open-book assessment task, so you may refer back to any of your mathematical resources.

You may find this formula useful: x =

Show full working steps throughout.

1. The diagram shows a photograph (shaded) mounted on cardboard. The cardboard measures 100 mm by 150 mm, and the border around the photograph is a uniform x mm wide. Which is the expression (in mm2) for the area of the photograph?

a. 150 × 100 – 150x – 100x b. 100 × 150 – 4x2

c. 100 × 150 – 4x d. (100 + 2x)(150 + 2x) e. (100 – 2x)(150 – 2x)

2. Solve the simultaneous equations: 3x + 2y = 11 2x – 3y = 16

3. Find algebraically, the point(s) of intersection of the line y = x + 6 and the curve y = x2 + 3x – 2.

4. When a pistol which fires flares is fired vertically on the moon, the height h of the flare above the surface is given by h = 2 + 60 t – 0.85 t2 metres where t is the time in seconds after firing the flare.

At what time does the flare reach a height of 50 metres as it rises above the surface of the moon?

5. Find the value of the discriminant and state the number of real root(s) of the quadratic equation x2 + 4x – 1 = 0.

–b ± b2 – 4ac2ac

150 mm

x

x

100

mm

6B


review activity

6. For each of the following,define suitable variable(s) and form appropriate equation(s). Do not solve the equations. a. I think of a number and decrease it by seven. The result is the same as dividing the original number by eight.

b. In a netball game Julie scored six more goals than Dianne. Together they scored 26 goals.

c. I’m thinking of two numbers. They differ by 19. Twice the greater number added to the smaller number is 59.

7. The length of a rectangular swimming pool is twice its width. The pool is surrounded by a concrete path one metre wide. If the area of the path is 46 square metres, find the dimensions of the swimming pool.

Show clearly the strategies you use to solve this problem.

8. An advertisement offers spring bulbs for sale. For $20, I could buy either 30 tulip bulbs and 20 daffodil bulbs, or 20 tulip bulbs and 40 daffodil bulbs.

Model this situation and use your model to find the price of one tulip bulb and the price of one daffodil bulb.

9. An airline collects a total of $11 858 for tickets from the 80 passengers on an airflight. Each child passenger pays $55 for their flight ticket. There are 20 children on the flight. Each adult passenger either buys a standard ticket for $198 or a thrifty ticket for $132 for the flight.

How many adult passengers have a thrifty ticket?

10. A rectangular pig pen has perimeter 39.8 metres and area 54.7 square metres. Find its dimensions.

Your teacher will assess this work.


answer guide

1. equation-solving techniques 1. 2(x – 5) = 5x + 2 2x – 10 = 5x + 2 2x – 5x = 2 + 10 –3x = 12 x = –4

3. = – 6

8 × = 8 × – 8 × 6

x = 4x – 48 x – 4x = –48 –3x = –48

x = 16

5. x² = 3x x² – 3x = 0 x(x – 3) = 0 Either x = 0 or x – 3 = 0 x = 3 x {0, 3}

7. x + 12 = x² x² – x – 12 = 0 (x – 4)(x + 3) = 0 Either x – 4 = 0 or x + 3 = 0 x = 4 x = –3 x {–3, 4}

1A

x8

x2

x8

x2

2323

7

2. 3(3x + 2) – 2(x – 3) = 5 9x + 6 – 2x + 6 = 5 7x + 12 = 5 7x = –7 x = –1

4. x² – 5x + 6 = 0 (x – 2)(x – 3) = 0 Either x – 2 = 0 or x – 3 = 0 x = 2 or x = 3 x {2, 3}

6. 3x² = 7x + 6 3x² – 7x – 6 = 0 (3x + 2)(x – 3) = 0 Either 3x + 2 = 0 or x – 3 = 0

x = − x = 3

x {– , 3}


answer guide

1B

1C

2. y = 2x + 6 (A) 2y + x = 7 (B)

Comparing (A) with y = mx + c, we have m = 2, c = 6. For (B), we complete a table of ordered pairs:

Solution is (–1, 4).

2. a. 7x – 2y = 3 (A) 3x + 8y = –43 (B) 28x – 8y = 12 (A) × 4 31x = –31 x = –1 –7 – 2y = 3 y = –5 Solution is (–1, –5).

b. 2x + y = 10 (A) 3x + 5y = 29 (B) –10x – 5y = –50 (A) × –5 –7x = –21 x = 3 6 + y = 10 y = 4 Solution is (3, 4).

x

y 3 2 1

1 3 5

1. y = 3x – 3 (A) y = –2x + 7 (B)

Comparing with y = mx + c, in (A), m = 3, c = –3 in (B), m = –2, c = 7.

Solution is (2, 3).

1. 4x – 3y = 7 (A) 2x + 3y = –1 (B) 6x = 6 x = 1 In (A), 4 – 3y = 7 y = –1 Solution is (1, –1).

0

y

x

y = 2x + 6

-2

-2-4 42 6

2

6

8

2y + x = 7

y

x2 40-2

y = 3x − 3

-2

2

4

6

y = –2x + 7


answer guide

2. c. 2x – 3y = 1 (A) 3x – 4y = 7 (B) 6x – 9y = 3 (A) × 3 –6x + 8y = –14 (B) × –2 –y = –11 y = 11 In (A): 2x−33 = 1 x = 17 Solution is (17, 11).

1. a. y = 4x + 1 y = x + 4 x + 4 = 4x + 1 x – 4x = 1 – 4 –3x = –3 x = 1 y = 1 + 4 y = 5 Solution is (1, 5).

2 a. 2x + 3y = 1 y = 2x – 5 2x + 3(2x – 5) = 1 2x + 6x – 15 = 1 8x = 16 x = 2 y = 2 × 2 – 5 y = –1 Solution is (2, –1).

1D

d. 3x + 2y = 13 (A) 7x + 3y = 27 (B) 9x + 6y = 39 (A) × 3 –14x – 6y = –54 (B) × –2 –5x = –15 x = 3 In (A): 9 + 2y = 13 y = 2 Solution is (3, 2).

b. y = 2x – 3 y = 4x + 5 4x + 5 = 2x – 3 4x – 2x = –3 – 5 2x = –8 x = –4 y = 4 × –4 + 5 y = –16 + 5 y = –11 Solution is (–4, –11).

b. 2x – 3y = 11 y = 3x – 6 2x – 3(3x – 6) = 11 2x – 9x + 18 = 11 –7x = –7 x = 1 y = 3 × 1 – 6 y = –3 Solution is (1, –3).


answer guide

c. 3x + 4y = 31 2x = y + 6 2x – 6 = y 3x + 4(2x – 6) = 31 3x + 8x – 24 = 31 11x = 55 x = 5 2 × 5 = y + 6 10 = y + 6 y = 4 Solution is (5, 4).

2. translating words into mathematics

1. a. 2x b. – x

c. d. x + 3

e. x – 5 f. 5 – x

g. 2x + 3 h. 2(x + 3)

i. j. + 3

1. a. z – 2 = 4

b. q + 6 = 10

c. m + 4 = 9

d. 7n = 21 or n × 7 = 21

e. 17 – x = 4

2A

2B

d. 6x – 3y = –8 x + 2y = 7 x = –2y + 7 6(–2y + 7) – 3y = –8 –12y + 42 – 3y = –8 –15y + 42 = –8 –15y = –50

y =

y = 3

x + 2 × 3 = 7

x + 6 = 7

x =

Solution is ( , 3 ).

2. a. x + 7 b. 7b

c. 6 – m d. 2g

e. 2x + 2 f. or a

g. 2h + 3 h. x – 4

i. or y j. 2p – 5

2. a. 3x + 7 = 42

b. (x – 5)² = 36

c. x + 6 = 3x

d. x + = 4.25

e. 2x – x² = –32

5015

13

13

13

13

1323

12

x4

x−52

x7

a2

12

y3

13

1x


answer guide

3. a. i. Let the numbers be x and y. ii. x + y = 42 x – y = 3

c. i. Let the numbers be u and v. ii. u + v = 4 + = –0.3

1. a. 2m b. m – 5

c. 2m + 3 d. m – 7

2. a. i. Let the remaining distance be x km. ii. 15 + x = 34 or 34 – x = 15

c. i. Let the number of children be c. ii. 3c + 3 = 24

e. i. Let the width of the rectangle be x metres.

ii. x + 18 + x + 18 = 56 or 2(x + 18) = 56 or 2x + 36 = 56

1. Let the width of the rectangle be x metres.

x + 2x + x + 2x = 96 6x = 96 x = 16

The lawn’s width is 16 m. Check 16 + 32 + 16 + 32 = 96

b. i. Let the numbers be m and n. ii. m + n = 27 mn = 50

b. i. Let the bottom mark be m. ii. 4 m = 48

d. i. Let the postal code be x. ii. x = 5 × 500 + 5

f. i. Let the number of sheep Charlie sold be x. ii. George sells x + 50 sheep x + x + 50 = 160

2. Letthefirstofthenumbersbem.m + (m + 1) + (m + 2) + (m + 3) + (m + 4) = 725 5m + 10 = 725 5m = 715 m = 143 Thefirstnumberis143. Check 143 + 144 + 145 + 146 + 147 = 725

1u

1v

2C

2D

18

x

2x

x


answer guide

3. Let the price of the cake be k cents. Price of pie is 2k Price of cola is k – 40 (k – 40) + k + 2k = 880 4k – 40 = 880 4k = 920 k = 230

Check Cost of pie, cake and cola = 2.30 + 4.60 + 1.90 = 8.80 The price of the cake was $2.30.

5. a. Let the present age of the younger tree be x years. i. So present age of older tree is 7x years. ii. Infiveyearsʼtimetheageof the younger tree is x + 5 years. iii.Infiveyearsʼtime,theageofthe older tree is 7x + 5 years. b. 7x + 5 = 4(x + 5) c. 7x + 5 = 4x + 20 7x – 4x = 20 – 5 3x = 15 x = 5 Theyoungertreeisfiveyearsoldnow. Check In5yearsʼtime,theyounger tree will be 10 years old. The older tree is now 35 years old,soinfiveyearsitwill be 40 years old. This is four times the age of the younger tree.

4. a. Letthespeedforthefirsttwohoursbex km/h b. Inthefirsttwohourshetravels2x km c. Speedforfinalthreehoursis(x – 15) km/h d. In these three hours he travels 3(x – 15) km e. 2x + 3(x – 15) = 480 f. 2x + 3x – 45 = 480 5x = 525 x = 105 g. The motorist travelled at 105 km/h for the first2hours. h. Check Travels210kminfirst2hours. Travels for 3 hours at 90 km/h; that is 270 km, so, total distance is 480 km.

6. Let the size of the smaller angle be mº. The other angle is (2m – 3)°

m + 2m – 3 = 180 3m = 183 m = 61 The smaller angle is 61°. Check The larger angle = 2 × 61 – 3 = 122 – 3 = 119° 119° + 61° = 180°

m° (2m−3)°


answer guide

7. Let the number of years JJ has belonged to the club be x. JJ has paid $336 subscription for x – 3 years.

Total = entrancefee + subforfirst3years + subfor spent remaining years 2 099 = 35 + 720 + 336(x – 3) 2 099 = 755 + 336x – 1 008 2 099 = –253 + 336x 2 352 = 336x 7 = x

JJ has belonged to the club for 10 years. Check Entrance fee 35 Sub for 3 years at $240 per year 720 Sub for 7 years at $192 per year 1 344 Total $2 099

3. quadratic equations

1. x² + 5x + 3 = 0 can be written as 1x² + 5x + 3 = 0 a = 1, b = 5, c = 3

x =

x =

x =

x =

x = or x =

x = –0.70 or x = –4.30 (2 d.p.)

3A

−b ± b²−4ac2a

−5 ± 5²−4×1×32 × 1

−5 ± 25−122

−5 ± 132

−5 + 132

−5 − 132

2. 4x² + 9x – 4 = 0 a = 4, b = 9, c = –4 x = Using the formula.

x =

x =

x = or x =

x = 0.38 or x = –2.63 (2 d.p.)

−9 ± 9²−4×4×−42 × 4

−9 ± 81 + 648

−9 ± 1458

−9 + 1458

−9 − 1458


answer guide

3. x² – 3x – 2 = 0 a = 1, b = –3, c = –2

x =

x =

x =

x =

x = or x =

x = 3.56 x = –0.56 (2 d.p.)

5. x² – 6x + 5 = 0 a = 1, b = –6, c = 5

x =

x =

x = But 16 is exactly 4

x =

Either

x = or x =

x = x =

x = 5 x = 1

Note: The solutions for this quadratic equation could have been found by factorising

x²−6x + 5 = (x−5)(x−1)

4. 7s² – 3s = 1 Rearranging 7s² – 3s – 1 = 0 a = 7, b = –3, c = –1

s =

s =

s =

s = or s =

s = 0.65 s = –0.22 (2 d.p.)

6. 4x² = –12x – 9 4x² + 12x + 9 = 0 a = 4, b = 12, c = 9

x =

x =

x =

x =

x = –1

−b ± b²−4ac2a

−(−3) ± (−3)²−4×1×−22 × 1

3 ± 9 + 82

3 ± 172

3 + 172

3 − 172

−(−3) ± (−3)²−4×7×−12 × 7

3 ± 9 + 2814

3 ± 3714

3 + 3714

3 − 3714

102

−(−6) ± (−6)²−4×1×52 × 1

6 ± 36−202

6 ± 162

6 ± 42

6 + 42

6 −42

22

−12 ± (12)²−4×4×92 × 4

−12 ± 144−1448

−12 ± 08

−128

12


answer guide

1. Since the discriminant is less than 0, the equation has no real roots.

3. The discriminant is greater than 0 and not the square of a rational number, so there are two irrational roots.

4. a. x² + 3x – 10 = 0 1x² + 3x + –10 = 0 a = 1, b = 3, c = –10 b² – 4ac = 3² – 4 × 1 × –10 = 9 + 40 = 49

b² – 4ac > 0 and b² – 4ac = 7² (7 Q).

There are two real roots, both of which are rational.

c. 4x² – 4x + 1 = 0 a = 4, b = –4, c = 1 b² – 4ac = (–4)² – 4 × 4 × 1 = 16 – 16 = 0 Since b² – 4ac = 0, there is one rational root.

e. 3x² – 2x + 3 = 0 a = 3, b = –2, c = 3 b² – 4ac = (–2)² – 4 × 3 × 3 = 4 – 36 = –32 Since b² – 4ac < 0, there are no real roots.

3B 2. Since b² – 4ac = 0, there is one repeated rational root.

b. x² + 3x – 11 = 0 1x² + 3x + –11 = 0 a = 1, b = 3, c = –11 b² – 4ac = 3² – 4 × 1 × –11 = 9 + 44 = 53

b² – 4ac > 0 and b² – 4ac is not the square of a rational number.

There are two real roots, both of which are irrational. d. x² + 2x + 1 = 0 a = 1, b = 2, c = 1 b² – 4ac = 2² – 4 × 1 × 1 = 0 Since b² – 4ac = 0, there is one rational root.

f. 2x² – 5x + 2 = 0 a = 2, b = –5, c = 2 b² – 4ac = (–5)² – 4 × 2 × 2 = 9 b² – 4ac > 0 and b² – 4ac = 3² (3 Q). There are two real roots, both of which are rational.


answer guide

5. The roots will be real if b² – 4ac≥0 x² – 6x + m = 0 a = 1, b = –6, c = m (–6)² – 4 × 1 × m ≥ 0 36 – 4 m ≥ 0 36≥ 4m 9≥ m The roots are real if m≤9.

6. 2x² + kx + 18 = 0 a = 2, b = k, c = 18 The roots are equal if b² – 4ac = 0 k² – 4 × 2 × 18 = 0 k² = 144 k = ±12 The roots are equal if k = ± 12.

1. x2 + 5x + 3 = 0 a = 1, b = 5, c = 3 b2 – 4ac = 52 – 4 × 1 × 3 = 13 b2 – 4ac>0→twodistinctsolutions

3. x2 – 3x – 2 = 0 a = 1, b = -3, c = -2 b2 – 4ac = (-3)2 – 4 × 1 × -2 = 17 b2 – 4ac>0→twodistinctsolutions

5. x2 – 6x + 5 =0 a = 1, b = - 6, c = 5 b2 – 4ac = (-6)2 – 4 × 1 × 5 = 16 b2 – 4ac>0→twodistinctsolutions

7. 3y2 = y – 1 3y2 – y + 1 = 0 a = 3, b = -1, c = 1 b2 – 4ac = (-1)2 – 4 × 3 × 1 = -11 b2 – 4ac<0→norealsolutions

9. Equation 7 has no real solutions

11. Equations 1 to 4 have irrational solutions [b2 – 4ac≠0orasquarenumber]

3C 2. 4x2 + 9x – 4 = 0 a = 4, b = 9, c = -4 b2 – 4ac = 92 – 4 × 4 × -4 = 145 b2 – 4ac>0→twodistinctsolutions

4. 7x2 – 3x = 1 7x2 – 3x – 1 = 0 a = 7, b = -3, c = -1 b2 – 4ac = (-3)2 – 4 × 7 × -1 = 37 b2 – 4ac>0→twodistinctsolutions

6. 4x2 + 12x + 9 = 0 a = 4, b = 12, c = 9 b2 – 4ac = 122 – 4 × 4 × 9 = 0 b2 – 4ac=0→twoequalsolutions

8. Equations 1 to 6 have real solutions

10. Equations 5 and 6 have rational solutions [b2 – 4ac=0orasquarenumber]


answer guide

discovery exercise1. a. x² – 4x + 3 = 0 a = 1, b = –4, c = 3 b² – 4ac = (–4)² – 4 × 1 × 3 = 16 – 12 = 4

c. The graph cuts the x-axis at (1, 0) and (3, 0).

b. f (x) = 0 (x – 1)(x – 3) = 0 x {1, 3}

d. f (x) = x² – 4x + 3 f (0) = 3 The graph cuts the y-axis at (0, 3). The axis of symmetry is x = 2 f (2) = 2² – 4 × 2 + 3 = –1 The vertex is (2, –1).

0 x

y

x = 2

f ( x ) = ( x 1 ) ( x 3 )

-2

2

2 4 6


answer guide

2. a. x² – 4x + 4 = 0 a = 1, b = –4, c = 4 b² – 4ac = (–4)² – 4 × 1 × 4 = 16 – 16 = 0

c. The graph touches the x-axis at (2, 0).

3. a. x² – 4x + 5 = 0 a = 1, b = –4, c = 5 b² – 4ac = (–4)² – 4 × 1 × 5 = 16 – 20 = –4

c. The graph does not intersect the x-axis.

b. g(x) = 0 (x – 2)² = 0 x {2}

d. g(x) = x² – 4x + 4 g(0) = 4 The graph cuts the y-axis at (0, 4). The axis of symmetry is x = 2 so the vertex is at (2, 0).

b. h(x) = 0 has no real solution since the discriminant is negative.

0 x

y

x = 2

g (x ) = ( x 2 )22

4

2 4


answer guide

4.

a. The roots of the equation ax² + bx + c = 0 are the x-values of the points where the graph of the function y = ax² + bx + c intersects the x-axis.

d. h(x) = x² – 4x + 5 h(0) = 5 The graph cuts the y-axis at (0, 5).

If h(x) = 5 5 = x² – 4x + 5 0 = x² – 4x 0 = x(x – 4) x {0, 4} A second point on the graph is (4, 5). The axis of symmetry is x = 2 h(2) = 2² – 4 × 2 + 5 = 1 The vertex is at (2, 1).

FunctionIntersection point(s)

on x-axis

Sign of b² = 4ac(+, 0, -)

Solutions of y = 0

Number ofreal rootsof y = 0

f(x)

g(x)

h(x)

(1, 0), (3, 0)

(2, 0)

none

+

0

-

x {1, 3}

x {2}

x { }

2

1

0

b. i. If the graph of y = ax² + bx + c crosses the x-axis twice, then ax² + bx + c = 0 has two real roots and b² – 4ac > 0.

ii. If the graph of y = ax² + bx + c touches the x-axis, then ax² + bx + c = 0 has one real root and b² – 4ac = 0.

iii. If the graph of y = ax² + bx + c does not intersect the x-axis, then ax² + bx + c = 0 has no real roots and b² – 4ac < 0.

4

6

x

y

x = 2

h(x) = x2 – 4x + 52

0 2 4 6


answer guide

1. a. The parabola cuts the x-axis at –1 and 3 so f(x) = 0 x {–1, 3}

2. a. The parabola does not cut the x-axis so the solution set is the empty set { } or Ø.

c. The discriminant has a negative sign.

multichoice

3. 3x2 – 4x + 2 = 0 a = 3, b = –4, c = 2 b2 – 4ac = (–4)2 – 4 × 3 × 2 = – 8 Answer: a. (the roots are not real)

4. problems involving quadratic equations

1. When the temperature is 300°C:

300 = 18 + 70x – 4x²

so 4x² – 70x + 282 = 0

a = 4, b = –70, c = 282

x = The quadratic formula.

x =

=

= 11.21 or 6.29 (2 d.p.)

Because 0 < x < 8, the value 11.21 canbeeliminated.Sothetimetofirst reach 300°C is 6.29 seconds (2 d.p.).

3D

4A

b. Since there are two real solutions, the discriminant is positive.

b. g(x) has no real roots.

4. If the equation has two real roots, then the parabola will cut the x-axis at two distinct points.

Answer: c (II and III)

2. When the ball is 18 m above the ground 18 = 2 + 20t – 5t² 0 = 5t² – 20t + 16 a = 5, b = –20, c = 16

t =

Using the quadratic formula.

t =

= 2.89 or 1.11 (2 d.p.)

The ball is 18 m above the ground at 1.11 seconds (2 d.p.) and 2.89 seconds (2 d.p.). There are two different times when the ball can be 18 metres above the ground as the ball can be rising or falling.

−b ± b²−4ac2a

−(−70) ± (−70)²−4×4×2822 × 4

70 ± 3888

−(−20) ± (−20)²−4×5×162 × 5

20 ± 8010


answer guide

3. Let the width of the path be x metres Area of outer rectangle = (16 + 2x)(25 + 2x) Area of inner rectangle = 16 × 25

(2x + 16)(2x + 25) – 16 × 25 = 180 4x² + 50x + 32x + 400 – 400 = 180 4x² + 82x – 180 = 0 2x² + 41x – 90 = 0

(2x + 45)(x – 2) = 0

Either 2x + 45 = 0 or x – 2 = 0 x = –22.5 or x = 2

Since x = –22.5 is not a realistic answer for a width, the width of the path is 2 metres. Check Area outer rectangle = 20 × 29 = 580 m² Area inner rectangle = 400 m² Area path = 580 – 400 = 180 m²

5. Let Kit’s age be k years. Kim’s age is k + 4 k² + (k + 4)² = 346 k² + k² + 8k + 16 = 346 2k² + 8k – 330 = 0 k² + 4k – 165 = 0 (k – 11)(k + 15) = 0 Either k – 11 = 0 or k + 15 = 0 k = 11 or k = –15 Since –15 is not a realistic age, Kit’s age is 11 years. Check If Kit is 11, Kim is 15 11² + 15² = 121 + 225 = 346

x

x

16

25

4. Let the son’s present age be n years. Father’s age is n + 26 years n(n + 26) = 192 n² + 26n – 192 = 0 (n + 32)(n – 6) = 0 Either n + 32 = 0 or n – 6 = 0 n = –32 or n = 6

Since n = –32 is not a realistic answer, the son’s present age is 6 years. Check If the son is now 6, his father is 32. The product of 6 and 32 is 192.


answer guide

6. a. (3x)² + (x – 2)² = (x + 7)²

c. If x = –1.45 then 2x and x – 2 have negative values, which are not realistic lengths for the sides. When x = 3.45 the sides 2x and x – 2 have positive values. Hence x = 3.45 is the correct solution.

5. problems involving simultaneous equations

1. Let the numbers be x and y. 3x + 2y = 30 (A) 2x + y = 17 (B) From (B): y = 17 – 2x In (A): 3x + 2(17 – 2x) = 30 3x + 34 – 4x = 30 –x = –4 x = 4

In (B): 2 × 4 + y = 17 y = 9 (x , y) = (4, 9)

Check In (A): 3 × 4 + 2 × 9 = 30

−b ± b²−4ac2a

−(−18) ± (−18)²−4×9×−452 × 9

18 ± 194418

b. 9x² + x² – 4x + 4 = x² + 14x + 49 9x² – 18x – 45 = 0 a = 9, b = –18, c = –45

x =

=

=

x = 3.45 or –1.45 (2 d.p.)

d. The sides are 3x = 10.35 cm (2 d.p.) x – 2 = 1.45 cm (2 d.p.) x + 7 = 10.45 cm (2 d.p.)

2. Let the greater number be x and the smaller number be y. x – y = 7 (A) 2x + 5y = 42 (B) From (A): x = 7 + y In (B): 2(7 + y) + 5y = 42 14 + 2y + 5y = 42 7y = 28 y = 4 In (A): x – 4 = 7 x = 11 (x, y) = (11, 4)

Check In (B): 2 × 11 + 5 × 4 = 42

1.45 cm

10.35 cm

10.45 cm

5A


answer guide

3. Let the two numbers be x and y.

(x + y) = – (A) (x – y) = (B)

Multiply both equations by 2 to remove fractions x + y = –1 (C) x – y = 3 (B) Add: 2x = 2 x = 1 In (C): 1 + y = –1 y = –2 (x, y) = (1, –2)

Check In (B): (1 – –2) = × 3

=

5. Let the number of boys be b and the number of girls be g. b + g = 37 (A) b = g + 9 (B) Substitute g + 9 for b in (A): g + 9 + g = 37 2g = 28 g = 14

In (B): b = 14 + 9 b = 23

There are 14 girls and 23 boys on the picnic.

Check 14 + 23 = 37 (people on picnic) 23 – 14 = 9 (9 more boys than girls)

1212

12

12

1232

32

4. Let the man’s age now be x years and his son’s age now be y years. x = y + 32 (A) x + y = 56 (B) Substitute y + 32 for x in (B): y + 32 + y = 56 2y = 24 y = 12 In (A): x = 12 + 32 = 44

The man’s present age is 44 and his son’s age is 12. Check Sum of ages = 44 + 12 = 56

Man’s age when son born = 44 – 12 = 32

6. Let Kelly’s weight be k kg and Terri’s weight be t kg. k = t – 3 (A) k + t + 5 = 90 (B)

Substitute t – 3 for k in (B) t – 3 + t + 5 = 90 2t = 88 t = 44

In (A): k = 44 – 3 = 41

Kelly weighs 41 kg and Terri weighs 44 kg.

Check Kelly weighs 3 kg less than Terri. Combined weight is 85 kg, which is 5 kg less than 90 kg.


answer guide

5B

7. Let the average price of a nectarine be n cents and the average price of a peach be p cents. 5n + 3p = 715 (A) 4n + 6p = 860 (B) –10n – 6p = –1 430 (A) × –2

Add: –6n = –570 n = 95

In (A): 5 × 95 + 3p = 715 3p = 240 p = 80

The average price of a nectarine is 90 cents and the average price of a peach is 80 cents.

Note: In the last two questions, the variablesmustbedefinedfortheweight and price, respectively. ‘Let Kelly be k’ or ‘The nectarine is n’arenotsufficiently welldefined.

1. y = x² + 3x – 2 (A) y = x – 3 (B) x – 3 = x² + 3x – 2 0 = x² + 2x + 1 0 = (x + 1)(x + 1) x = –1 If x – 1, then y = –1 – 3 = –4

Check in (A): (–1)² + 3 × –1 – 2 = 1 – 3 – 2 = –4 Solution is: (x, y) {(–1, –4)}

2. x² + y² = 10 (A) y = 3x + 10 (B) x² + (3x + 10)² = 10 x² + 9x² + 60x + 100 = 10 10x² + 60x + 90 = 0 x² + 6x + 9 = 0 (x + 3)² = 0 x = –3

If x = –3, then y = 3 × –3 + 10 = 1

Check in (A): (–3)² + 1² = 10 Solution: (x, y) {(–3, 1)}.


answer guide

3. y = x² + x – 5 (A) 2x – y + 1 = 0 (B) From (B): y = 2x + 1 2x + 1 = x² + x – 5 0 = x² – x – 6 0 = (x – 3)(x + 2) Either x – 3 = 0 or x + 2 = 0 x = 3 x = –2

If x = 3, y = 2 × 3 + 1 = 7 If x = –2, y = 2 × –2 + 1 = –3 Check In (A): (3, 7) gives 3² + 3 – 5 = 7 In (A): (–2, –3) gives (–2)2 + –2 – 5 = –3 Solution: (x, y) {(3, 7), (–2, –3)}.

5. xy = 15 (A) 4x – y = 7 (B) From (B): y = 4x – 7 x(4x – 7) = 15 4x² – 7x – 15 = 0 (4x + 5)(x – 3) = 0

Either 4x + 5 = 0 or x – 3 = 0

x = – x = 3

If x = – , y = 4 × – – 7 = –12

If x = 3, y = 4 × 3 – 7 = 5

Check in (A) – × –12 = 15

3 × 5 = 15 ✓

Solution is: (x, y) {(– , –12), (3, 5)}.

4. x² + y² = 25 (A) x + y = 7 (B) From (B): y = 7 – x x² + (7 – x)² = 25 x² + 49 – 14x + x² = 25 2x² – 14x + 24 = 0 x² – 7x + 12 = 0 (x – 3)(x – 4) = 0 Either x – 3 = 0 or x – 4 = 0 x = 3 x = 4

If x = 3, y = 7 – 3 = 4 If x = 4, y = 7 – 4 = 3 Check in (A) 3² + 4² = 9 + 16 = 25 ✓ Solution is: (x, y) {(3, 4), (4, 3)}.

54

54

54

54

54


answer guide

1. Let the numbers be x and y. • The sum is 25 x + y = 25 (A) • The product is 144 xy = 144 (B) From (A): y = 25 – x x(25 – x) = 144 25x – x² = 144 0 = x² – 25x + 144 0 = (x – 16)(x – 9) Either x – 16 = 0 or x−9=0 x = 16 or x = 9 If x = 16, y = 9 If x = 9, y = 16 The numbers are 9 and 16. Check 16 × 9 = 144

Solution is: (x, y) {(16, 9), (9, 16)}.

3. Let the dimensions be x centimetres and y centimetres.

• The area is 325 cm² xy = 325 (A)• The perimeter is 76 cm 2x + 2y = 76 (B) From (B): 2x = 76 – 2y x = 38 – y y(38 – y) = 325 38y – y² = 325 0 = y² – 38y + 325 0 = (y – 13)(y – 25) Either y – 13 = 0 or y – 25 = 0 y = 13 y = 25

If y = 13, x = 25 If y = 25, x = 13

Check Area = 25 × 13 = 325 Perimeter = 2 × 13 + 2 × 25 = 76 The dimensions of the rectangle are 25 centimetres by 13 centimetres.

2. Let the numbers be x and y. • The difference is 5 x – y = 5 (A) • Sum of squares is 97 x² + y² = 97 (B) From (A): x = y + 5 (y + 5)² + y² = 97 y² + 10y + 25 + y² = 97 2y² + 10y – 72 = 0 y² + 5y – 36 = 0 (y + 9)(y – 4) = 0 Either y + 9 = 0 or y – 4 = 0 y = –9 or y = 4 If y = –9, x = –4 If y = 4, x = 9

Since the numbers are positive, the numbers are 9 and 4. Check 9 – 4 = 5 9² + 42 = 81 + 16 = 97 Solution is: (x, y) {(–4, –9), (9, 4)}.

5C

x

y


answer guide

4. Let the lengths of the sides be x centimetres and y centimetres.

• The perimeter is 120 cm x + y + 51 = 120 (A)• The hypotenuse is 51 x² + y² = 51² (B) From (A): x = 69 – y (69 – y)² + y² = 512

4761 – 138y + 2y² = 2 601 2y² – 138y + 2 160 = 0 y² – 69y + 1 080 = 0 (y – 24)(y – 45) = 0

Either y – 24 = 0 or y – 45 = 0 y = 24 y = 45

If y = 24, x = 45 If y = 45, x = 24

Check In (B): 24² + 45² = 576 + 2 025 = 2 601 = 51²

The lengths of the two other sides are 24 centimetres and 45 centimetres.

1. a. i. y = (x + 3)(1 – x) y = 3 ii. (0, 3), (–2, 3) b. i. x² + y² = 25 y = –2x + 5 ii. (0, 5), (4, –3)

y

x

51 cm

5D


answer guide

6A

2. a. Here is the algebraic solution for graph (a). y = (x + 3)(1 – x) (A) y = 3 (B) 3 = (x + 3)(1 – x) 3 = –x2 – 2x + 3 x² + 2x = 0 x(x + 2) = 0 Either x = 0 or x + 2 = 0 x = –2

Solution is: (x, y) {(0, 3), (–2, 3)}.

b. Here is the algebraic solution for graph (b). x² + y² = 25 (A) y = –2x + 5 (B) x² + (–2x + 5)² = 25 x² + 4x² – 20x + 25 = 25 5x² – 20x = 0 5x(x – 4) = 0

Either 5x = 0 or x – 4 = 0 x = 0 or x = 4

If x = 0, then y = –2 × 0 + 5 = 5 If x = 4, then y = –2 × 4 + 5 = –3 Solution is: (x, y) {(0, 5), (4, –3)}.

b. 2x + 5y = 34 (A) 3x + 4y = 30 (B) –8x – 20y = –136 (A) × –4 15x + 20y = 150 (B) × 5 7x = 14 x = 2 2 × 2 + 5y = 34 5y = 30 y = 6 Solution is (2, 6).

6. review activity

1. a. x – 2y = 7 y = 2x + 22 x – 2(2x + 22) = 7 x – 4x – 44 = 7 –3x = 51 x = –17 y = 2 × –17 + 22 y = –12

Solution is (–17, –12).


answer guide

3. Let the smaller integer be x The next even integer is x + 2 2x – 18 = x + 2 x = 20 The two integers are 20 and 22.

Check 2 × 20 – 18 = 40 – 18 = 22

5. Let the length of the shorter side be x cm

x² + (x + 15)2 = 34² Pythagoras

x² + x² + 30x + 225 = 1 156

2x² + 30x – 931 = 0

a = 2, b = 30, c = −931

x =

x =

= 15.34 or –30.34 (2 d.p.)

Since a negative length is unrealistic, the value of x is 15.34 only. Hence the lengths of the two sides are: x = 15.3 cm (1 d.p.) and x + 15 = 15.34 + 15 = 30.3 cm (1 d.p.)

2. You must state clearly what each letter represents. a. Let x be the number 2 + 5x b. Let x be the number 2x – 6

c. Let the two numbers be x and y

d. Let t be the number of hours 60t

4. Let the number of adults’ tickets sold be a. Let the number of children’s tickets sold be c. a + c = 300 (A) 5a + 3c = 1 080 (B) Add: −3a – 3c = −900 2a = 180

In (A): 90 + c = 300 c = 210

210 children’s and 90 adults’ tickets were sold.

Check Number of tickets sold = 90 + 210 = 300

Proceeds from sales = 90 × 5 + 210 × 3 = $1 080

x

x + 15

34 cm

−b ± b²−4ac2a

−30 ± 30²−4×2×−9312 × 2

x + y2


answer guide

6. x² + y² = 4 (A) y = x – 2 (B) x² + (x – 2)² = 4 x² + (x² – 4x + 4) = 4 x² + x² – 4x = 0 2x² – 4x = 0 x² – 2x = 0 x(x – 2) = 0 Either x = 0 or x – 2 = 0 x = 2 If x = 0, y = 0 – 2 y = –2 If x = 2, y = 2 – 2 y = 0 Check In (A) (0, –2) gives 0² + (–2)² = 4 In (A) (2, 0) gives 2² + 0² = 4

Solution is: (x, y) {(0, –2), (2, 0)}.

8. x² + y² = 9 (A) x + 3 = 0 (B) From (B): x = –3 Substitute –3 for x in (A): (–3)² + y² = 9 y² = 0 y = 0

Solution is: (−3, 0).

Since the line intersects the circle at one point only, the line x = –3 is a tangent.

7. y = –5x + 1 (A) xy = –6 (B) Substitute –5x + 1 for y in (B): x(–5x + 1) = –6 –5x² + x + 6 = 0 5x² – x – 6 = 0 (5x – 6)(x + 1) = 0 Either 5x – 6 = 0 or x + 1 = 0

x = x = –1

If x = , y = –5 × + 1 = –5 If x = –1, y = –5 × –1 + 1 = 6

Check In (B): ( , –5) gives × –5 = –6

In (B): (–1, 6) gives –1 × 6 = –6

The points of intersection are ( , –5) and (–1, 6)

9. a. y = x² y = 12 – x

b. (x, y) {(–4, 16), (3, 9)}

c. x² = 12 – x x² + x – 12 = 0 The solutions to this equation are x {−4, 3}. They are the x-coordinates of the points of intersection.

65

65

65

65

65


answer guide

10. Let the lengths of the sides be x centimetres and y centimetres.

• The wire is 56 cm long x + y + 25 = 56 (A)• It forms a right-angled triangle x² + y² = 25² (B) From (A): x = 31 – y In (B): (31 – y)² + y² = 25² 961 – 62y + y² + y2 = 625 2y² – 62y + 336 = 0 y² – 31y + 168 = 0 (y – 7)(y – 24) = 0 Either y – 7 = 0 or y – 24 = 0 y = 7 y = 24

If y = 7, x = 31 – 7 x = 24 If y = 24, x = 31 – 24 x = 7 Check In (B): 24² + 7² = 576 + 49 = 625 = 25²

The lengths of the other two sides of the triangle are 24 centimetres and 7 centimetres.

25 cm

x

y


appendix

derivation of the quadratic formula

Consider the general equation: ax² + bx + c = 0

Divide through by a to make x² + x + = 0thecoefficientof x² equal to 1.

Move the constant term to the other side. x² + x =

Complete the square for

x² + x by adding ( × )² x² + x + ( )² = + ( )²

that is, ( )² to both sides.

Write the left-hand side as a perfect square (x + )² = + and simplify the square on the right.

Add the fractions on the right-hand side. (x + )² = +

(x + )² = +

(x + )² =

Take the square root of both sides. x + = ±

Simplify the square root using the rule x + =

=

x + =

Move to the right-hand side. x = ±

Write the result as a single fraction. x =

ba

ca

ba

−ca

ba

12

ba

ba

b2a

−ca

b2a

b2a

b2a

−ca

b²4a²

b2a

−c × 4aa × 4a

b²4a²

b2a

−4ac4a²

b²4a²

b2a

b²−4ac4a²

b2a

b²−4ac4a²

b2a

± b²−4ac4a²

pq

pq

b2a

± b²−4ac2a

b2a

−b2a

b²−4ac2a

−b ± b²−4ac2a


acknowledgements

Every effort has been made to acknowledge and contact copyright holders. Te Aho o Te Kura Pounamu apologises for any

omissions and welcomes more accurate information.

Graph diagrams created using GeoGebra program from http://www.geogebra.org/cms. Used in any medium for education

and its promotion by permission.

All other illustrations copyrighted to Te Aho o Te Kura Pounamu, Wellington, NZ.

Photos:

istock: Solving problems – 15798783, School picnic – 14466457,

Theatre – 15377330, Peaches – 16017618, Tangerines – 16643212.

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mathematics and statistics - Te Kura

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